Mathematics, 03.08.2019 06:10 brmoen1392

Aquality analyst of a tennis racquet manufacturing plant investigates if the length of a junior's tennis racquet conforms to the standards. he randomly selects twenty racquets and obtains the following lengths: 21, 25, 23, 22, 24, 21, 25, 21, 23, 26, 21, 24, 22, 24, 23, 21, 21, 26, 23, 24 assume that lengths of tennis racquets are normally distributed. construct a 99.9% confidence interval for the mean length of all the junior's tennis racquets manufactured at this plant.

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Answer from: jackiesantos1298

Confidence interval : 21.506 to 24.493

Step-by-step explanation:

A quality analyst selects twenty racquets and obtains the following lengths:

21, 25, 23, 22, 24, 21, 25, 21, 23, 26, 21, 24, 22, 24, 23, 21, 21, 26, 23, 24

So, sample size = n =20

Now we are supposed to find Construct a 99.9% confidence interval for the mean length of all the junior's tennis racquets manufactured at this plant.

Since n < 30

So we will use t-distribution

Confidence level = 99.9%

Significance level = α = 0.001

Now calculate the sample mean

X=21, 25, 23, 22, 24, 21, 25, 21, 23, 26, 21, 24, 22, 24, 23, 21, 21, 26, 23, 24

Sample mean = $\bar{x}=\frac{\sum x}{n}$

Sample mean = $\bar{x}=\frac{21+25+23+22+24+21+25+21+23+ 26+ 21+24+22+ 24+23+21+ 21+ 26+23+ 24}{20}$

Sample mean = $\bar{x}=23$

Sample standard deviation = $\sqrt{\frac{\sum(x-\bar{x})^2}{n-1}}$

Sample standard deviation = $\sqrt{\frac{(21-23)^2+(25-23)^2+(23-23)^2+(22-23)^2+(24-23)^2+(21-23)^2+(25-23)^2+(21-23)^2+(23-23)^2+(26-23)^2+(21-23)^2+(24-23)^2+(22-23)^2+(24-23)^2+(23-23)^2+(21-23)^2+(21-23)^2+(26-23)^2+(23-23)^2+(24-23)^2}{20-1}}$

Sample standard deviation= s = 1.72

Degree of freedom = n-1 = 20-1 -19

Critical value of t using the t-distribution table $t_{\frac{\alpha}{2}$ = 3.883

Formula of confidence interval : $\bar{x} \pm t_{\frac{\alpha}{2}} \times \frac{s}{\sqrt{n}}$

Substitute the values in the formula

Confidence interval : $23 \pm 1.73 \times \frac{1.72}{\sqrt{20}}$

Confidence interval : $23 -3.883 \times \frac{1.72}{\sqrt{20}}$ to $23 + 3.883 \times \frac{1.72}{\sqrt{20}}$

Confidence interval : 21.506 to 24.493

Hence Confidence interval : 21.506 to 24.493

Answer from: itsbrizee

The 99.9% confidence interval for the mean length of all the junior's tennis racquets manufactured at this plant is [21.7958,24.2042].

Step-by-step explanation:

The given data set is

21, 25, 23, 22, 24, 21, 25, 21, 23, 26, 21, 24, 22, 24, 23, 21, 21, 26, 23, 24

Mean of the data set is

$Mean=\frac{\sum x}{n}$

$\overline{x}=\frac{21+25+23+22+24+21+25+21+23+26+21+24+22+24+23+21+21+26+23+24}{20}$

$\overline{x}=23$

Sample mean of the data is 23.

Standard deviation of population is

$\sigma=\sqrt{\frac{\sum (x-\overline{x})^2}{n}}$

$\sigma=\sqrt{\frac{56}{20}}$

$\sigma\approx 1.673$

Assume that lengths of tennis racquets are normally distributed.

The value of z is 3.219 at 99.9% confidence interval .

We need to find the 99.9% confidence interval for the mean length of all the junior's tennis racquets manufactured at this plant.

$Interval=\overline{x}\pm z*\cdot \frac{\sigma}{\sqrt{n}}$

$Interval=23\pm 3.219\cdot \frac{1.673}{\sqrt{20}}$

$Interval=23\pm 1.2042$

Interval=[23-1.2042,23+1.2042]

Interval=[21.7958,24.2042]

Therefore the 99.9% confidence interval for the mean length of all the junior's tennis racquets manufactured at this plant is [21.7958,24.2042].

Answer from: Quest

Ithink the answer is: it is likely.

Answer from: Quest

answer:

step-by-step explanation:

here, the total dishes = 10

dishes having meat = 3 ( chicken, meatloaf and hamburger )

thus, the probability of having meat on the chosen dishes = number of dishes having meat / total dishes

therefore, the probability of not having meat = 1- the probability of having meat on the chosen dishes

⇒ option b is correct.

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